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and 
be side lengths of a triangle with perimeter 
. Show that![\[a^2+b^2+c^2+abc<8.\]](http://latex.artofproblemsolving.com/d/9/8/d98a071538c07301938da20cf2a293b0936d848c.png)
b) Is there a real number 
such that for all triangles with perimeter we have![\[a^2+b^2+c^2+abc<d \quad\]](http://latex.artofproblemsolving.com/c/d/5/cd5e002c21106833f3aa3fb08b67a60d0d5eb10e.png)
where and are the side lengths of the triangle?
be a triangle with incentre 
and circumcircle 
. Let 
and 
be the second intersection points of with 
and 
, respectively. The chord 
meets 
at a point 
, and 
at a point 
. Let 
be the intersection point of the line through parallel to 
and the line through parallel to 
. Suppose that the tangents to at 
and 
meet at a point 
. Prove that the three lines 
and 
are either parallel or concurrent.
is written on the board. Every step, we replace the written number 
into the number 
, where 
is a prime divisor of , and this step continues for infinity. Prove that the number 
was chosen as the number infinitely many times.
be the circumcenter of triangle . The altitudes from 
of triangle intersects the circumcircle of the triangle at 
respectively. 
meets 
at 
respectively. Prove that the circumcircles of triangles 
share two common points.
such that for any two infinite sequences of positive integers 
and 
, such that 
for all 
, there always exists a real number such that 
for all 
.
such that for any two infinite sequences of positive integers and , such that for all , there always exists a real number such that for all .
and 
of any length are considered. The answer for will be slightly different.
. It is relatively straightforward to show that all other do not work so we take care of that first.
consider the sequences 
and 
. Now, say there exists a real number which satisfies the desired conditions. Let 
where 
and 
. Then,![\[m=\lfloor (m+\epsilon) \rfloor \equiv 1 \pmod{n}\]](http://latex.artofproblemsolving.com/6/d/c/6dc8676d1b0e27d06dc6d532b7fa7e9dc940e817.png)
Further,![\[2m+\lfloor 2\epsilon \rfloor = \lfloor 2(m+\epsilon)\rfloor \equiv 1 \pmod{n}\]](http://latex.artofproblemsolving.com/e/f/2/ef2e5ff90f0f603bed5f6d8a851f0853892c9811.png)
which since 
indicates,![\[\lfloor 2\epsilon \rfloor \equiv -1 \pmod{n}\]](http://latex.artofproblemsolving.com/3/9/0/390d72910492aa4589e2f0baffb44f3955c0e438.png)
Now, since we must have 
so 
which is a contradiction or 
which is also a contradiction for . Thus, there exists no such for this case as claimed.. For any pair of sequences and we show that we can match the terms 
for a sufficiently large number of terms which implies the result. This is because, if there exists a pair of sequences and for which such does not exist, there must exist some real number 
for which the above equality holds for the maximum first consecutive terms. But if we can provide which matches this for an arbitrary number of leading terms this is a contradiction. We handle this via induction.
. When 
, 
and since 
implies 
. If 
then let 
(thus 
) and if 
then let 
(and thus 
). Now, say there exists a non-empty range for 
which matches the condition for all 
and 
for some 
.
. Let 
. Since 
and 
we must have 
. So, 
(which includes atleast two non-negative integers). Now, select 
such that 
and let 
so 
which satisfies all the conditions of the inductive hypothesis and completes the induction.
and we show that we can match the terms for a sufficiently large number of terms which implies the result.
such that 
for all 
for all 
, but does not exist when can be unbounded.
,
.
,![\[\lfloor 2c\rfloor\equiv 1\pmod n\]](http://latex.artofproblemsolving.com/5/7/b/57bdbd10b6b9a77edf8c6a026f995b4acdc86073.png)
![\[\lfloor{4c}\rfloor\equiv 1\pmod n\]](http://latex.artofproblemsolving.com/5/3/3/533a5ca2b3609513078b264da52dd4d70643ee76.png)
contradiction as 
or 
.
.
contains a 
after the point, let 
be the first time it occurs. Then,![\[\lfloor 2^k\{c\}\rfloor\equiv 0\pmod 2\]](http://latex.artofproblemsolving.com/2/9/0/290f9f1f284c79c37e19a505294b727b8fa75d90.png)
![\[\lfloor a_k c\rfloor\equiv 0\pmod 2\]](http://latex.artofproblemsolving.com/c/3/b/c3b71d0a1ca562d7101dc9fc604afcf845333d24.png)
contradiction. Otherwise, 
,
.
for all 
.![\[ \lfloor{ca_i}\rfloor\equiv b_i\pmod 2\]](http://latex.artofproblemsolving.com/c/9/f/c9f51bdb9e11d5647f6f0f691b896d6e83cc1721.png)
is equivalent to there exists 
such that![\[ \frac{2x_i+b_i}{a_i}\le c < \frac{2x_i+b_i+1}{a_i} \]](http://latex.artofproblemsolving.com/2/7/3/2734a7f8c2445d96bbf417e957e36656d4ca9a53.png)
Similarly, the ceiling condition is equivalent to![\[ \frac{2x_i+b_i}{a_i}< c \le \frac{2x_i+b_i+1}{a_i} \]](http://latex.artofproblemsolving.com/3/8/e/38e7c2db2972a69bc8e46a95689a907a31343ad8.png)
We will now show that there exists 
such that![\[ \frac{2x_1+b_1}{a_1}\le \frac{2x_2+b_2}{a_2}\le \cdots\]](http://latex.artofproblemsolving.com/5/b/d/5bdf9ebd97a6b68a612daf80d4d5dc1089613199.png)
and![\[ \frac{2x_1+b_1+1}{a_1}\ge \frac{2x_2+b_2+1}{a_2}\ge \cdots\]](http://latex.artofproblemsolving.com/4/e/d/4ed10f44de9ae3c5726bb82a967d65bdf90b5483.png)
Induct. If 
exists, then we find 
.![\[ (2x_i+b_i)d_{i+1} \le 2x_{i+1}+b_{i+1} \le (2x_i+b_i+1)d_{i+1}-1 \]](http://latex.artofproblemsolving.com/3/e/0/3e0d29acd659ff7c3eca93aed8492cae323fada7.png)
But 
so there exists at least two integers within the interval. One of them must be 
so 
is not eventually constant, take 
is not eventually constant, take 
strictly decreasing. 
and 
where 
represents fractional part
= 1